Question

1. $5x^3 + 2x - 8$
2. $100x^2 + 3$
3. $-9x^4 + 8x^3 - 7x + 1$

Explanation:

It seems the problem is about analyzing or perhaps factoring these polynomial expressions. Let's assume we are to determine if they can be factored (maybe over integers or reals, or just identify their nature). Let's take each one:

Problem 24: \(5x^3 + 2x - 8\)

Step 1: Try Rational Root Theorem

The Rational Root Theorem states that possible rational roots are factors of the constant term over factors of the leading coefficient. Here, constant term is \(-8\), leading coefficient is \(5\). So possible rational roots are \(\pm1, \pm2, \pm4, \pm8, \pm\frac{1}{5}, \pm\frac{2}{5}, \pm\frac{4}{5}, \pm\frac{8}{5}\). Let's test \(x = 1\): \(5(1)^3 + 2(1) - 8 = 5 + 2 - 8 = -1
eq 0\). \(x = \frac{4}{5}\): \(5(\frac{4}{5})^3 + 2(\frac{4}{5}) - 8 = 5(\frac{64}{125}) + \frac{8}{5} - 8 = \frac{64}{25} + \frac{40}{25} - \frac{200}{25} = \frac{104 - 200}{25} = -\frac{96}{25}
eq 0\). \(x = \frac{8}{5}\): \(5(\frac{8}{5})^3 + 2(\frac{8}{5}) - 8 = 5(\frac{512}{125}) + \frac{16}{5} - 8 = \frac{512}{25} + \frac{80}{25} - \frac{200}{25} = \frac{592 - 200}{25} = \frac{392}{25}
eq 0\). \(x = 2\): \(5(8) + 4 - 8 = 40 + 4 - 8 = 36
eq 0\). So no rational roots, and since it's a cubic, it might not factor over integers. Maybe over reals, but that's more complex. Alternatively, maybe it's already in simplest form.

Step 2: Conclusion

The polynomial \(5x^3 + 2x - 8\) does not have obvious integer factors (from Rational Root Theorem test) and is a cubic polynomial. So it may be irreducible over integers, or could be factored using more advanced methods (like Cardano's formula for cubics), but for basic factoring, it's likely considered as is or with complex methods.

Problem 25: \(100x^2 + 3\)

Step 1: Check for difference of squares or other factoring

A difference of squares is \(a^2 - b^2 = (a - b)(a + b)\). Here, we have \(100x^2 + 3 = (10x)^2 + (\sqrt{3})^2\). In real numbers, this is a sum of squares, which doesn't factor (since \(a^2 + b^2\) doesn't factor over reals into linear terms). Over complex numbers, it would be \((10x - i\sqrt{3})(10x + i\sqrt{3})\), but in integer or real factoring (for basic algebra), it's irreducible.

Step 2: Conclusion

The polynomial \(100x^2 + 3\) is a sum of squares and doesn't factor over the integers or real numbers (in basic factoring), so it's irreducible in that context.

Problem 26: \(-9x^4 + 8x^3 - 7x + 1\)

Step 1: Try Rational Root Theorem

Leading coefficient is \(-9\), constant term is \(1\). Possible rational roots are \(\pm1, \pm\frac{1}{3}, \pm\frac{1}{9}\). Test \(x = 1\): \(-9 + 8 - 7 + 1 = -7
eq 0\). \(x = -1\): \(-9(-1)^4 + 8(-1)^3 - 7(-1) + 1 = -9 - 8 + 7 + 1 = -9
eq 0\). \(x = \frac{1}{3}\): \(-9(\frac{1}{3})^4 + 8(\frac{1}{3})^3 - 7(\frac{1}{3}) + 1 = -9(\frac{1}{81}) + 8(\frac{1}{27}) - \frac{7}{3} + 1 = -\frac{1}{9} + \frac{8}{27} - \frac{63}{27} + \frac{27}{27} = (-\frac{3}{27} + \frac{8}{27} - \frac{63}{27} + \frac{27}{27}) = \frac{-3 + 8 - 63 + 27}{27} = \frac{-31}{27}
eq 0\). \(x = \frac{1}{9}\): This would be very small, let's compute: \(-9(\frac{1}{9})^4 + 8(\frac{1}{9})^3 - 7(\frac{1}{9}) + 1 = -9(\frac{1}{6561}) + 8(\frac{1}{729}) - \frac{7}{9} + 1 = -\frac{1}{729} + \frac{8}{729} - \frac{567}{729} + \frac{729}{729} = \frac{-1 + 8 - 567 + 729}{729} = \frac{169}{729}
eq 0\). So no rational roots.

Step 2: Check for quadratic factors

Assume it factors as \((ax^2 + bx + c)(dx^2 + ex + f)\). Multiply out: \(adx^4 + (ae + bd)x^3 + (af + be + cd)x^2 + (bf + ce)x + cf\). Compare with \(-9x^4 + 8x^3 + 0x^2 - 7x + 1\) (note the \(x^2\) term is 0). So:

• \(ad = -9\) (possible pairs: \(a= -9, d=1\); \(a=9, d=-1\); \(a=3, d=-3\); etc.)
• \(ae + bd = 8\)
• \(af + be + cd = 0\)
• \(bf + ce = -7\)
• \(cf = 1\) (so \(c = f = 1\) or \(c = f = -1\))

Let's try \(c = 1, f = 1\). Then:

From \(ad = -9\), let's take \(a = -9, d = 1\). Then:

\(ae + bd = -9e + b(1) = 8\) --> \(b - 9e = 8\)

\(af + be + cd = -9(1) + b e + (-9)(1) = -9 + be - 9 = be - 18 = 0\) --> \(be = 18\)

\(bf + ce = b(1) + (-9)(1) = b - 9 = -7\) --> \(b = 2\)

If \(b = 2\), then from \(be = 18\), \(2e = 18\) --> \(e = 9\). Then from \(b - 9e = 8\), \(2 - 9(9) = 2 - 81 = -79
eq 8\). Not good.

Try \(a = 9, d = -1\), \(c = 1, f = 1\):

\(ae + bd = 9e + b(-1) = 8\) --> \(9e - b = 8\)

\(af + be + cd = 9(1) + b e + 9(1) = 9 + be + 9 = be + 18 = 0\) --> \(be = -18\)

\(bf + ce = b(1) + 9(1) = b + 9 = -7\) --> \(b = -16\)

Then from \(be = -18\), \(-16e = -18\) --> \(e = \frac{18}{16} = \frac{9}{8}\). Not integer. Then \(9e - b = 9(\frac{9}{8}) - (-16) = \frac{81}{8} + 16 = \frac{81 + 128}{8} = \frac{209}{8}
eq 8\).

Try \(c = -1, f = -1\):

\(af + be + cd = a(-1) + b e + d(-1) = -a + be - d = 0\)

\(bf + ce = b(-1) + a(-1) = -b - a = -7\) --> \(a + b = 7\)

\(ad = -9\), \(cf = (-1)(-1) = 1\)

Let \(a = -9, d = 1\):

\(ae + bd = -9e + b(1) = 8\) --> \(b - 9e = 8\)

\(-a + be - d = 9 + be - 1 = be + 8 = 0\) --> \(be = -8\)

\(a + b = -9 + b = 7\) --> \(b = 16\)

Then \(be = 16e = -8\) --> \(e = -\frac{1}{2}\). Then \(b - 9e = 16 - 9(-\frac{1}{2}) = 16 + \frac{9}{2} = \frac{41}{2}
eq 8\). Not good.

This is getting complex. Alternatively, since there are no rational roots and factoring into quadratics is difficult, this polynomial is likely irreducible over the integers, or requires more advanced factoring (like over reals or complex, but for basic algebra, it's irreducible).

Answer:

s:

1. The polynomial \(5x^3 + 2x - 8\) does not factor over the integers (by Rational Root Theorem test) and is a cubic. It may be irreducible in basic factoring.
2. \(100x^2 + 3\) is a sum of squares and does not factor over the integers or real numbers (in basic factoring), so it's irreducible.
3. The polynomial \(-9x^4 + 8x^3 - 7x + 1\) has no rational roots (by Rational Root Theorem) and factoring into quadratics is difficult, so it's likely irreducible over the integers in basic factoring.

(If the question was to factor them, and assuming we are to state if they factor or not, these are the conclusions. If more specific, like factor over complex, that's different, but for basic algebra, these are the results.)